Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{5x - 5}{5x^2 + 5x - 10} \div \dfrac{3x^2 + 9x}{x^3 - 4x^2 - 21x} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{5x - 5}{5x^2 + 5x - 10} \times \dfrac{x^3 - 4x^2 - 21x}{3x^2 + 9x} $ First factor out any common factors. $z = \dfrac{5(x - 1)}{5(x^2 + x - 2)} \times \dfrac{x(x^2 - 4x - 21)}{3x(x + 3)} $ Then factor the quadratic expressions. $z = \dfrac {5(x - 1)} {5(x - 1)(x + 2)} \times \dfrac {x(x + 3)(x - 7)} {3x(x + 3)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {5(x - 1) \times x(x + 3)(x - 7) } { 5(x - 1)(x + 2) \times 3x(x + 3)} $ $z = \dfrac {5x(x + 3)(x - 7)(x - 1)} {15x(x - 1)(x + 2)(x + 3)} $ Notice that $(x - 1)$ and $(x + 3)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {5x(x + 3)(x - 7)\cancel{(x - 1)}} {15x\cancel{(x - 1)}(x + 2)(x + 3)} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $z = \dfrac {5x\cancel{(x + 3)}(x - 7)\cancel{(x - 1)}} {15x\cancel{(x - 1)}(x + 2)\cancel{(x + 3)}} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $z = \dfrac {5x(x - 7)} {15x(x + 2)} $ $ z = \dfrac{x - 7}{3(x + 2)}; x \neq 1; x \neq -3 $